Over which fields is the (appropriate version of the) "Sylvester law of inertia" valid?

Long version:

Let $V$ be a finite dimensional vector space over the field $\Bbbk$ of characteristic different from $2$ (so quadratic forms are the same as symmetric bilinear forms). Consider the "discriminant" function

$$\mathrm{discr}: \mathrm{Sym}^2(V^*)\to \Bbbk/\Bbbk^2,$$

where $\Bbbk/\Bbbk^2$ is the quotient of the multiplicative monoid $\Bbbk$ by the squares, which maps a (possibly degenerate) symmetric bilinear form $q$ to the determinant of it's matrix relative to any base, which is well defined up to multiplication by squares. Of course, this is not a complete invariant, as different degenerate quadratic forms all have discriminant $0$.

Given $q\in\mathrm{Sym}^2(V^*)$, and an orthogonal basis $B$ for $V$ in which $q$ has the form $\Sigma_i\lambda_ix_iy_i$, we have a "signature" map

$$\sigma_q^B:\Bbbk/\Bbbk^2\to\mathbb{N}$$

sending a $\lambda$ in the domain to the number of vectors in $B$ for wich the corresponding $\lambda_i$ has image $\lambda$ in $\Bbbk/\Bbbk^2$.

1) Is $\sigma_q^B$ independent from $B$ ?

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2) If the answer to 1) is "yes", is $q\mapsto\sigma_q$ a complete invariant for quadratic forms over $\Bbbk$ ? That is, is it true that two quadratic spaces $(V,q)$ and $(V',q')$ are isomorphic if and only if $\sigma_q=\sigma_{q'}$ ?

For $\Bbbk=\mathbb{R}$ or $\mathbb{C}$ the answer to 1) and 2) is "yes": it's is the usual Sylvester law of inertia. Maybe the above general question has a well known or even elementary answer?...

1 Answer
1

OK, I now have a complete answer; I'll delete the other shortly. The answer to question 2 is yes and easily so; I wonder if I am missing something. If one quadratic form is $\sum a_i x_i^2$, and other is $\sum b_i y_i^2$, and $a_i = b_i u_i^2$ for $u_i \neq 0$, then clearly the two quadratic forms are equivalent by the invertible change of variable $x_i = y_i u_i$.

Now for the interesting part. Let $K$ be a field. The following are equivalent:

(1) The answer to Question 1 is "yes".

(2) For every $a \in K$ we have (Condition 2a) either $a$ or $-a$ is square and (Condition 2b) the sum of two squares is a square.

(3) One of the following two cases occurs: (Case 3a) Every element of $K$ is square or (Case 3b) $K$ is an ordered field in which every positive element is square.

$(2) \implies (3)$: If $-1$ is square, then Condition 2a implies that everything is square. If not, define a total order on $K$ by $x \leq y$ if $y-x$ is square. Condition 2a implies that this is a total order; condition 2b implies that it is transitive and compatible with addition; I leave it as an exercise that it is compatible with multiplication by positive elements. So $K$ is an ordered field where the positive elements are the squares, and we are in condition (3b).

$(3) \implies (1)$. If we are in case 3a, then the statement is trivial because $K/K^2$ only contains one nonzero element. In case 3b, look up your favorite proof that signature is a well-defined invariant of real quadratic forms, and note that it only uses these facts.

I suspect some classical algebraist assigned a name to fields as in 3b. They are more special than formally real field (where $-1$ is not a sum of squares), but less special than real closed fields (where, in addition to the properties in 3b, we require that every odd degree polynomial have a root).

An example of a field as in 3b is the field of real numbers which have straight-edge and compass constructions.